Count of numbers up to N having only 4 factors or divisors

Given an integer N, find the number of natural numbers less than or equal to N and have 4 factors.

Example: 

Input: N = 8
Output: 2
Explanation: {1} is divisor set of 1
{1, 2} is divisor set of 2
{1, 3} is divisor set of 3
{1, 2, 4} is divisor set of 4
{1, 5} is divisor set of 5
{1, 2, 3, 6} is divisor set of 6
{1, 7} is divisor set of 7
{1, 2, 4, 8} is divisor set of 8
So, 6 and 8 are only natural numbers less than or equal to N and count of divisors 4.

Input: N = 2
Output: 0

Naive Approach:

The naive approach for the problem is that we can traverse from 1 to N and count number of divisors of each one of them. If the count comes out to be 4 , then increase the resultant number say res by 1. Finally at the end of traversal we will be having the number of natural numbers less than or equal to N and have 4 factors in the variable res. 

We will be counting divisors efficiently using the approach of complexity O(n^1/3) as mentioned and discussed in this article: Count Divisors of n in O(n^1/3).

Algorithm:

1. Define a function SieveOfEratosthenes that accepts n, a boolean array prime[], a boolean array primesquare[], and an integer array a[] as parameters.
2. Initialize all entries in prime[] as true and all entries in primesquare[] as false.
3. Mark prime[1] as false.
4. Use the Sieve of Eratosthenes algorithm to mark all non-prime numbers in the range 2 to n in the prime[] array.
5. Store all prime numbers in the range 2 to n in the integer array a[].
6. For each prime number p in the integer array a[], update the primesquare[p*p] to true.
7. Define a function countDivisors that accepts an integer n as a parameter.
8. If n is 1, return 1 as it will have only 1 as a factor.
9. Declare a boolean array prime[] and primesquare[] of size n+1 and n*n+1 respectively.
10. Declare an integer array a[] of size n.
11. Call the SieveOfEratosthenes function with parameters n, prime[], primesquare[], and a[].
12. Initialize a variable ans to 1 that will contain the total number of distinct divisors.
13. Loop through all prime numbers a[i] in the integer array a[] until a[i]^3 > n.
14. Calculate the power of a[i] in n using a while loop, incrementing the power for each iteration until n is no longer divisible by a[i].
15. Multiply ans by (cnt+1) where cnt is the power of a[i] in n.
16. If n is greater than 1 after the loop, then it has a prime factor greater than cube root of n, which will have a power of 1. Multiply ans by 2 in this case.
17. If n is a square of a prime number, multiply ans by 3.
18. If n is not equal to 1 after the above cases, it means n has two distinct prime factors. Multiply ans by 4 in this case.
19. Return ans.
20. Define the main function.
21. Initialize an integer N to the maximum range.
22. Initialize an integer variable res to 0 that will store the result.
23. Loop through each number from 1 to N.
24. If the count of divisors of the current number is 4, increment res by 1.
25. Print the value of res as the result.

Below is the implementation of the approach:

C++

// C++ code for the approach

#include <bits/stdc++.h>
using namespace std;

void SieveOfEratosthenes(int n, bool prime[],
                         bool primesquare[], int a[])
{

    //For more details check out: https://www.geeksforgeeks.org/sieve-of-eratosthenes/

    // Create a boolean array "prime[0..n]" and
    // initialize all entries it as true. A value
    // in prime[i] will finally be false if i is
    // Not a prime, else true.
    for (int i = 2; i <= n; i++)
        prime[i] = true;

    // Create a boolean array "primesquare[0..n*n+1]"
    // and initialize all entries it as false. A value
    // in squareprime[i] will finally be true if i is
    // square of prime, else false.
    for (int i = 0; i <= (n * n + 1); i++)
        primesquare[i] = false;

    // 1 is not a prime number
    prime[1] = false;

    for (int p = 2; p * p <= n; p++) {
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {
            // Update all multiples of p starting from p * p
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }

    int j = 0;
    for (int p = 2; p <= n; p++) {
        if (prime[p]) {
            // Storing primes in an array
            a[j] = p;

            // Update value in primesquare[p*p],
            // if p is prime.
            primesquare[p * p] = true;
            j++;
        }
    }
}

// Function to count divisors
int countDivisors(int n) {
    // If number is 1, then it will have only 1
    // as a factor. So, total factors will be 1.
    if (n == 1)
        return 1;

    bool prime[n + 1], primesquare[n * n + 1];

    int a[n]; // for storing primes upto n

    // Calling SieveOfEratosthenes to store prime
    // factors of n and to store square of prime
    // factors of n
    SieveOfEratosthenes(n, prime, primesquare, a);

    // ans will contain total number of distinct
    // divisors
    int ans = 1;

    // Loop for counting factors of n
    for (int i = 0;; i++) {
        // a[i] is not less than cube root n
        if (a[i] * a[i] * a[i] > n)
            break;

        // Calculating power of a[i] in n.
        int cnt = 1; // cnt is power of prime a[i] in n.
        while (n % a[i] == 0) // if a[i] is a factor of n
        {
            n = n / a[i];
            cnt = cnt + 1; // incrementing power
        }

        // Calculating the number of divisors
        // If n = a^p * b^q then total divisors of n
        // are (p+1)*(q+1)
        ans = ans * cnt;
    }

    // if a[i] is greater than cube root of n

    // First case
    if (prime[n])
        ans = ans * 2;

    // Second case
    else if (primesquare[n])
        ans = ans * 3;

    // Third case
    else if (n != 1)
        ans = ans * 4;

    return ans; // Total divisors
}

// Driver code
int main() {
      // Input
    int N = 8;

      // resultant counter
      int res = 0;

      // traversing through each number from 1 to N
    for(int i = 1; i <= N; i++) {
      // if count of divisors is 4
      // increase res by 1
      if( countDivisors(i) == 4)
        res += 1;
    }

      // Print res
      cout<< res << endl;
    return 0;
}

Java

import java.util.Arrays;

class Solution {

    static boolean[] prime;
    static boolean[] primesquare;

    public static int[] sieveOfEratosthenes(int n)
    {
        // Initialize a boolean array to mark prime numbers
        prime = new boolean[n + 1];
        Arrays.fill(prime, true);

        // Initialize a boolean array to mark square prime
        // numbers
        primesquare = new boolean[(n * n) + 1];
        Arrays.fill(primesquare, false);

        // Initialize an array to store prime numbers
        int[] a = new int[n];

        // 1 is not a prime number
        prime[1] = false;

        // Find prime numbers using the Sieve of
        // Eratosthenes algorithm
        int p = 2;
        while (p * p <= n) {
            if (prime[p]) {
                for (int i = p * p; i <= n; i += p) {
                    prime[i]
                        = false; // Mark multiples of prime
                                 // numbers as non-prime
                }
            }
            p++;
        }

        // Store prime numbers in the array
        int j = 0;
        for (p = 2; p <= n; p++) {
            if (prime[p]) {
                a[j] = p;
                primesquare[p * p]
                    = true; // Mark square prime numbers
                j++;
            }
        }

        return a;
    }

    public static int countDivisors(int n)
    {
        if (n == 1) {
            return 1;
        }

        // Get prime numbers and related information
        int[] a = sieveOfEratosthenes(n);
        int ans = 1;
        int i = 0;

        // Count divisors using the prime factorization of
        // the number
        while (true) {
            if (a[i] * a[i] * a[i] > n) {
                break;
            }

            int cnt = 1;
            while (n % a[i] == 0) {
                n = n / a[i];
                cnt++;
            }

            ans *= cnt;
            i++;
        }

        // Handle remaining factors
        if (n > 1) {
            if (prime[n]) {
                ans *= 2;
            }
            else if (primesquare[n]) {
                ans *= 3;
            }
            else {
                ans *= 4;
            }
        }

        return ans; // Return the total number of divisors
    }

    public static void main(String[] args)
    {
        int N = 8;
        int res = 0;

        // Count the number of numbers with exactly four
        // divisors
        for (int i = 1; i <= N; i++) {
            if (countDivisors(i) == 4) {
                res++;
            }
        }

        System.out.println(res); // Print the result
    }
}

Python3

# Function to find prime numbers using the Sieve of Eratosthenes algorithm
def SieveOfEratosthenes(n):
    prime = [True] * (n + 1)  # Initialize a boolean array to mark prime numbers
    primesquare = [False] * ((n * n) + 1)  # Initialize a boolean array to mark square prime numbers
    a = [0] * n  # Initialize an array to store prime numbers

    prime[1] = False  # 1 is not a prime number

    p = 2
    while p * p <= n:
        if prime[p]:
            for i in range(p * p, n + 1, p):
                prime[i] = False  # Mark multiples of prime numbers as non-prime
        p += 1

    j = 0
    for p in range(2, n + 1):
        if prime[p]:
            a[j] = p  # Store prime numbers in the array
            primesquare[p * p] = True  # Mark square prime numbers
            j += 1

    return a, prime, primesquare  # Return the arrays of prime numbers and related information


# Function to count divisors of a number
def countDivisors(n):
    if n == 1:
        return 1

    a, prime, primesquare = SieveOfEratosthenes(n)  # Get prime numbers and related information
    ans = 1
    i = 0

    while True:
        if a[i] * a[i] * a[i] > n:
            break
        cnt = 1
        while n % a[i] == 0:
            n = n // a[i]
            cnt += 1
        ans *= cnt

        i += 1

    if prime[n]:
        ans *= 2
    elif primesquare[n]:
        ans *= 3
    elif n != 1:
        ans *= 4

    return ans  # Return the total number of divisors


# Driver code
def main():
    N = 8
    res = 0
    for i in range(1, N + 1):
        if countDivisors(i) == 4:
            res += 1

    print(res)  # Print the result

if __name__ == "__main__":
    main()  # Call the main function if the script is executed directly

C#

using System;

class GFG
{
    static bool[] prime;
    static bool[] primesquare;

    public static int[] SieveOfEratosthenes(int n)
    {
        // Initialize a boolean array to mark prime numbers
        prime = new bool[n + 1];
        Array.Fill(prime, true);

        // Initialize a boolean array to mark square prime numbers
        primesquare = new bool[(n * n) + 1];
        Array.Fill(primesquare, false);

        // Initialize an array to store prime numbers
        int[] a = new int[n];

        // 1 is not a prime number
        prime[1] = false;

        // Find prime numbers using the Sieve of Eratosthenes algorithm
        int p = 2;
        while (p * p <= n)
        {
            if (prime[p])
            {
                for (int i = p * p; i <= n; i += p)
                {
                    prime[i] = false; // Mark multiples of prime numbers as non-prime
                }
            }
            p++;
        }

        // Store prime numbers in the array
        int j = 0;
        for (p = 2; p <= n; p++)
        {
            if (prime[p])
            {
                a[j] = p;
                primesquare[p * p] = true; // Mark square prime numbers
                j++;
            }
        }

        return a;
    }

    public static int CountDivisors(int n)
    {
        if (n == 1)
        {
            return 1;
        }

        // Get prime numbers and related information
        int[] a = SieveOfEratosthenes(n);
        int ans = 1;
        int i = 0;

        // Count divisors using the prime factorization of the number
        while (true)
        {
            if (a[i] * a[i] * a[i] > n)
            {
                break;
            }

            int cnt = 1;
            while (n % a[i] == 0)
            {
                n = n / a[i];
                cnt++;
            }

            ans *= cnt;
            i++;
        }

        // Handle remaining factors
        if (n > 1)
        {
            if (prime[n])
            {
                ans *= 2;
            }
            else if (primesquare[n])
            {
                ans *= 3;
            }
            else
            {
                ans *= 4;
            }
        }

        return ans; // Return the total number of divisors
    }

    public static void Main()
    {
        int N = 8;
        int res = 0;

        // Count the number of numbers with exactly four divisors
        for (int i = 1; i <= N; i++)
        {
            if (CountDivisors(i) == 4)
            {
                res++;
            }
        }

        Console.WriteLine(res); // Print the result
    }
}

Javascript

let primesquare;

function sieveOfEratosthenes(n)
{
    // Initialize a boolean array to mark prime numbers
    let prime = new Array(n + 1).fill(true);

    // Initialize a boolean array to mark square prime
    // numbers
    primesquare = new Array((n * n) + 1).fill(false);

    // Initialize an array to store prime numbers
    let a = [];

    // 1 is not a prime number
    prime[1] = false;

    // Find prime numbers using the Sieve of Eratosthenes
    // algorithm
    let p = 2;
    while (p * p <= n) {
        if (prime[p]) {
            for (let i = p * p; i <= n; i += p) {
                prime[i] = false; // Mark multiples of prime
                                  // numbers as non-prime
            }
        }
        p++;
    }

    // Store prime numbers in the array
    let j = 0;
    for (p = 2; p <= n; p++) {
        if (prime[p]) {
            a[j] = p;
            primesquare[p * p]
                = true; // Mark square prime numbers
            j++;
        }
    }

    return a;
}

function countDivisors(n)
{
    if (n == 1) {
        return 1;
    }

    // Get prime numbers and related information
    let a = sieveOfEratosthenes(n);
    let ans = 1;
    let i = 0;

    // Count divisors using the prime factorization of the
    // number
    while (true) {
        if (a[i] * a[i] * a[i] > n) {
            break;
        }

        let cnt = 1;
        while (n % a[i] == 0) {
            n = n / a[i];
            cnt++;
        }

        ans *= cnt;
        i++;
    }

    // Handle remaining factors
    if (n > 1) {
        if (a.includes(n)) {
            ans *= 2;
        }
        else if (primesquare[n]) {
            ans *= 3;
        }
        else {
            ans *= 4;
        }
    }

    return ans; // Return the total number of divisors
}

let N = 8;
let res = 0;

// Count the number of numbers with exactly four divisors
for (let i = 1; i <= N; i++) {
    if (countDivisors(i) == 4) {
        res++;
    }
}

console.log(res); // Print the result

Output

2

Time Complexity: O(N * N^1/3) as for loop which is running contributes O(N) and inside it countDivisors function contributes N^1/3 time. So, overall time complexity becomes O(N * N^1/3).  Here, N is the input integer.

Auxiliary Space: O(N) as arrays like prime and primesquare has been created. Here, N is the input integer.

Approach: The idea to solve the problem is based on the following observation:

• Any number M can be written in form M = p₁^e1 * p₂^e2 *  . . . where (p₁, p₂ . . .) are primes and (e₁, e₂ . . .) are respective exponents.
• The total number of factors of M is therefore (e + 1)*(e + 1)* . . .
• From above points, for the count of divisors of a natural number to be 4, there are two cases:-
  • Case-1: N = p1 * p2 (where p1 and p2 are two distinct prime numbers)
  • Case-2: N = p³ (where p is a prime number)
• So there must be two primes whose multiplication is less than N or one prime whose cube is less than N.

Follow the steps mentioned below to solve the problem:

• Find all the prime numbers less than or equal to N using the sieve of Eratosthenes.
• For Case-1 iterate through all prime numbers and use binary search to find a number of primes whose product is at most N.
• For Case-2, do a binary search to find the number of primes whose cube is less than or equal to N.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find primes <= N
vector<int> SieveOfEratosthenes(int n)
{
    // Create a boolean array
    // "prime[0..n]" and initialize
    // all entries it as true.
    // A value in prime[i] will
    // finally be false if i is
    // Not a prime, else true.
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));

    for (long long int p = 2;
         p * p <= n; p++) {

        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
            // Update all multiples
            // of p greater than or
            // equal to the square of it
            for (long long int i = p * p;
                 i <= n; i += p)
                prime[i] = false;
        }
    }

    // Vector for storing prime number
    // less than or equal to N
    vector<int> primes;

    // Store all prime numbers
    for (int p = 2; p <= n; p++)
        if (prime[p])
            primes.push_back(p);

    return primes;
}

// Find floor of cube root of N
int primeCubic(vector<int>& primes, int N)
{

    // Val stores cube root of N
    long long int l = 0, r = N, mid, val;

    // Binary search loop for finding
    // floor of cube root of N
    while (l <= r) {
        mid = (l + r) / 2;
        if ((mid * mid * mid) <= N) {
            val = mid;
            l = mid + 1;
        }
        else {
            r = mid - 1;
        }
    }

    // Iterator for finding index with
    // value just greater than Val in primes
    auto it = upper_bound(primes.begin(),
                          primes.end(), val);
    it--;

    return (it - primes.begin() + 1);
}

// Function to find primes with product <= N
int primeProduct(vector<int>& primes,
                 int N)
{
    // Stores the answer
    int answer = 0;

    // Iterator storing pointer to
    // current prime
    auto cur = primes.begin();

    // Loop for traversing all primes
    // Find number of indices less than
    // current indices for which product
    // is less than or equal to N
    for (auto i : primes) {
        long long int add
            = upper_bound(primes.begin(),
                          cur, (N / i))
              - primes.begin();
        answer += add;
        cur++;
    }
    return answer;
}

// Function to find the total count
int print(int N)
{
    vector<int> primes
        = SieveOfEratosthenes(N);

    int answer = 0;
    answer += primeCubic(primes, N);
    answer += primeProduct(primes, N);
    return answer;
}

// Driver code
int main()
{
    int N = 8;

    // Print function Call
    cout << print(N);
    return 0;
}

Java

// Java program for above approach
import java.util.*;

public class Solution
{

  // Function to find primes <= N
  static ArrayList<Integer> SieveOfEratosthenes(int n)
  {

    // Create a boolean array
    // "prime[0..n]" and initialize
    // all entries it as true.
    // A value in prime[i] will
    // finally be false if i is
    // Not a prime, else true.
    boolean[] prime = new boolean[n + 1];
    Arrays.fill(prime, true);

    for (int p = 2; p * p <= n; p++) {

      // If prime[p] is not changed,
      // then it is a prime
      if (prime[p] == true) {
        // Update all multiples
        // of p greater than or
        // equal to the square of it
        for (int i = p * p; i <= n; i += p)
          prime[i] = false;
      }
    }

    // Vector for storing prime number
    // less than or equal to N
    ArrayList<Integer> primes = new ArrayList<>();

    // Store all prime numbers
    for (int p = 2; p <= n; p++)
      if (prime[p])
        primes.add(p);

    return primes;
  }

  static int upper_bound(ArrayList<Integer> arr, int lo,
                         int hi, int key)
  {
    int mid, N = arr.size();

    // Initialise starting index and
    // ending index
    int low = lo;
    int high = hi;

    // Till low is less than high
    while (low < high && low != N) {
      mid = low + (high - low) / 2;

      // If key is greater than or equal
      // to arr[mid], then find in
      // right subarray
      if (key >= arr.get(mid)) {
        low = mid + 1;
      }

      // If key is less than arr[mid]
      // then find in left subarray
      else {
        high = mid;
      }
    }
    return low;
  }
  // Find floor of cube root of N
  static int primeCubic(ArrayList<Integer> primes, int N)
  {

    // Val stores cube root of N
    int l = 0, r = N, mid, val = 0;

    // Binary search loop for finding
    // floor of cube root of N
    while (l <= r) {
      mid = (l + r) / 2;
      if ((mid * mid * mid) <= N) {
        val = mid;
        l = mid + 1;
      }
      else {
        r = mid - 1;
      }
    }

    // Iterator for finding index with
    // value just greater than Val in primes
    int it = upper_bound(primes, 0, primes.size(), val);
    it--;

    return (it + 1);
  }

  // Function to find primes with product <= N
  static int primeProduct(ArrayList<Integer> primes,
                          int N)
  {
    // Stores the answer
    int answer = 0;

    // Iterator storing pointer to
    // current prime
    int cur = 0;

    // Loop for traversing all primes
    // Find number of indices less than
    // current indices for which product
    // is less than or equal to N
    for (int i : primes) {
      int add = upper_bound(primes, 0, cur, (N / i));
      answer += add;
      cur++;
    }
    return answer;
  }

  // Function to find the total count
  static int print(int N)
  {
    ArrayList<Integer> primes = SieveOfEratosthenes(N);

    int answer = 0;
    answer += primeCubic(primes, N);
    answer += primeProduct(primes, N);
    return answer;
  }

  // Driver code
  public static void main(String[] args)
  {
    int N = 8;

    // Print function Call
    System.out.println(print(N));
  }
}

// This code is contributed by karandeep1234.

Python3

# Python3 program for the above approach
import bisect

# Function to find primes <= N
def SieveOfEratosthenes(n):

    # Create a boolean array
    # "prime[0..n]" and initialize
    # all entries it as true.
    # A value in prime[i] will
    # finally be false if i is
    # Not a prime, else true.
    prime = [True] * (n + 1)

    for p in range(2, 1 + int(n ** 0.5)):

        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == True):
            # Update all multiples
            # of p greater than or
            # equal to the square of it
            for i in range(p * p, n + 1, p):
                prime[i] = False

    # Vector for storing prime number
    # less than or equal to N
    primes = []

    # Store all prime numbers
    for p in range(2, n + 1):
        if prime[p]:
            primes.append(p)


    return primes


# Find floor of cube root of N
def primeCubic(primes, N):

    #Val stores cube root of N
    l = 0
    r = N

    # Binary search loop for finding
    # floor of cube root of N
    while (l <= r):
        mid = (l + r) // 2
        if ((mid * mid * mid) <= N):
            val = mid
            l = mid + 1

        else:
            r = mid - 1

    # Iterator for finding index with
    # value just greater than Val in primes
    it = bisect.bisect_right(primes, val)

    return it

# Function to find primes with product <= N
def primeProduct(primes, N):

    # Stores the answer
    answer = 0

    # Iterator storing pointer to
    # current prime
    cur = 0

    # Loop for traversing all primes
    # Find number of indices less than
    # current indices for which product
    # is less than or equal to N
    for i in primes:
        add = bisect.bisect_right(primes[:cur], N // i)
        answer += add
        cur += 1

    return answer

# Function to find the total count
def print_(N):

    primes = SieveOfEratosthenes(N)

    answer = 0
    answer += primeCubic(primes, N)
    answer += primeProduct(primes, N)
    return answer


# Driver code
N = 8

# Print function Call
print(print_(N))

# This code is contributed by phasing17

C#

using System;
using System.Collections.Generic;
class GFG
{

  // Function to find primes <= N
  static int[] SieveOfEratosthenes(int n)
  {

    // Create a boolean array
    // "prime[0..n]" and initialize
    // all entries it as true.
    // A value in prime[i] will
    // finally be false if i is
    // Not a prime, else true.
    bool[] prime = new bool[n + 1];
    for (int i = 0; i <= n; i++) {
      prime[i] = true;
    }

    for (int p = 2; p * p <= n; p++) {

      // If prime[p] is not changed,
      // then it is a prime
      if (prime[p] == true) {
        // Update all multiples
        // of p greater than or
        // equal to the square of it
        for (int i = p * p; i <= n; i += p)
          prime[i] = false;
      }
    }

    // Array for storing prime number
    // less than or equal to N
    // int []primes= new int[];
    List<int> primes = new List<int>();

    // Store all prime numbers
    for (int p = 2; p <= n; p++)
      if (prime[p])
        primes.Add(p);

    return primes.ToArray();
  }

  static int upper_bound(int[] arr, int N, int X)
  {
    int mid;
    int low = 0;
    int high = N;
    while (low < high) {
      mid = low + (high - low) / 2;
      if (X >= arr[mid])
        low = mid + 1;
      else
        high = mid;
    }
    if (low < N && arr[low] <= X)
      low++;
    return low;
  }

  // Find floor of cube root of N
  static int primeCubic(int[] primes, int N)
  {

    // Val stores cube root of N
    int l = 0, r = N, mid, val = 0;

    // Binary search loop for finding
    // floor of cube root of N
    while (l <= r) {
      mid = (l + r) / 2;
      if ((mid * mid * mid) <= N) {
        val = mid;
        l = mid + 1;
      }
      else {
        r = mid - 1;
      }
    }

    // Iterator for finding index with
    // value just greater than Val in primes
    int it = upper_bound(primes, primes.Length, val);
    return it;
  }

  // Function to find primes with product <= N
  static int primeProduct(int[] primes, int N)
  {

    // Stores the answer
    int answer = 0;

    // Iterator storing pointer to
    // current prime
    int cur = 0;

    // Loop for traversing all primes
    // Find number of indices less than
    // current indices for which product
    // is less than or equal to N
    for (int i = 0; i < primes.Length; i++) {
      int add
        = upper_bound(primes, cur, (N / primes[i]));
      answer += add;
      cur++;
    }
    return answer;
  }

  // Function to find the total count
  static int print(int N)
  {
    int[] primes = SieveOfEratosthenes(N);

    int answer = 0;
    answer += primeCubic(primes, N);
    answer += primeProduct(primes, N);
    return answer;
  }
  static void Main()
  {
    int N = 8;

    // Print function Call
    Console.Write(print(N));
  }
}

// This code is contributed by garg28harsh.

Javascript

// Function to find primes <=N
const SieveOfEratosthenes = (n) => {

    // Create a boolean array
    // "prime[0..n]" and initialize
    // all entries it as true.
    // A value in prime[i] will
    // finally be false if i is
    // Not a prime, else true.
  let prime = Array(n + 1).fill(true);
  for (let p = 2; p * p <= n; p++)
  {

      // If prime[p] is not changed,
      // then it is a prime
    if (prime[p])
    {

      // Update all multiples
      // of p greater than or
      // equal to the square of it
      for (let i = p * p; i <= n; i += p)
        prime[i] = false;
    }
  }

// Vector for storing prime number
// less than or equal to N
  let primes = [];

// Store all prime numbers
  for (let p = 2; p <= n; p++)
    if (prime[p])
      primes.push(p);

  return primes;
}

const upper_bound = (arr, lo, hi, key) => {
  let mid, N = arr.length;

// Initialise starting index and
// ending index
  let low = lo;
  let high = hi;

// Till low is less than high
  while (low < high && low != N) {
    mid = low + Math.floor((high - low) / 2);

    // If key is greater than or equal
      // to arr[mid], then find in
      // right subarray
    if (key >= arr[mid]) {
      low = mid + 1;
    }

    // If key is less than arr[mid]
    // then find in left subarray
    else {
      high = mid;
    }
  }
  return low;
}

// Find floor of cube root of N
const primeCubic = (primes, N) => {

  // Val stores cube root of N
  let l = 0, r = N, mid, val = 0;

 // Binary search loop for finding
    // floor of cube root of N
  while (l <= r) {
    mid = Math.floor((l + r) / 2);
    if ((mid * mid * mid) <= N) {
      val = mid;
      l = mid + 1;
    }
    else {
      r = mid - 1;
    }
  }

    // Iterator for finding index with
    // value just greater than Val in primes
  let it = upper_bound(primes, 0, primes.length, val);
  it--;

  return (it + 1);
}

 // Function to find primes with product <= N
const primeProduct = (primes, N) => {
  let answer = 0;

    // Iterator storing pointer to
    // current prime
  let cur = 0;

   // Loop for traversing all primes
    // Find number of indices less than
    // current indices for which product
    // is less than or equal to N
  for (let i of primes) {
    let add = upper_bound(primes, 0, cur, Math.floor(N / i));
    answer += add;
    cur++;
  }
  return answer;
}

// Function to find the total count
const print = (N) => {
  let primes = SieveOfEratosthenes(N);

  let answer = 0;
  answer += primeCubic(primes, N);
  answer += primeProduct(primes, N);
  return answer;
}

console.log(print(8));

// This code is contributed by anskalyan3.

Output

2

Time Complexity: O(N * log(logN) + N + logN) ≈ O(N * log (logN))
Auxiliary Space: O(N)